To tackle this problem, we need to analyze the system of a solid conducting sphere and a concentric spherical conducting shell. Let's break it down step by step for clarity.
Understanding the Setup
We have a solid conducting sphere with a radius of 10 cm, and it is surrounded by a thick spherical conducting shell. The inner radius of the shell is 15 cm, and the outer radius is 20 cm. The space between the inner sphere and the outer shell is filled with air, which acts as a dielectric medium.
Part A: Finding the Potential Difference and Capacitance
When a charge of 5 nC is placed on the inner conductor, we can find the potential difference between the two conductors and the capacitance of the system.
1. Capacitance Calculation
The capacitance \( C \) of a spherical capacitor formed by two concentric spheres is given by the formula:
C = 4 \pi \epsilon_0 \frac{r_1 r_2}{r_2 - r_1}
Where:
- \( r_1 \) = radius of the inner sphere = 0.1 m (10 cm)
- \( r_2 \) = radius of the outer shell = 0.2 m (20 cm)
- \( \epsilon_0 \) = permittivity of free space = \( 8.85 \times 10^{-12} \, \text{F/m} \)
Substituting the values:
C = 4 \pi (8.85 \times 10^{-12}) \frac{(0.1)(0.2)}{0.2 - 0.1}
Calculating this gives:
C = 4 \pi (8.85 \times 10^{-12}) \frac{0.02}{0.1} = 4 \pi (8.85 \times 10^{-12}) \times 0.2
After calculating, we find:
C \approx 5.57 \times 10^{-12} \, \text{F} \, (or \, 5.57 \, pF)
2. Potential Difference Calculation
The potential \( V \) of a charged conductor is given by:
V = \frac{Q}{C}
Where \( Q \) is the charge on the inner sphere (5 nC = \( 5 \times 10^{-9} \, C \)). Thus:
V_{inner} = \frac{5 \times 10^{-9}}{5.57 \times 10^{-12}} \approx 899.64 \, V
Now, the potential at the outer shell (which is grounded) is 0 V. Therefore, the potential difference \( V_{diff} \) between the inner and outer conductors is:
V_{diff} = V_{inner} - V_{outer} = 899.64 \, V - 0 \, V = 899.64 \, V
Part B: Finding Charges on Spheres with Given Potentials
Now, we need to determine the charges on the spheres when the potentials are given as 90 V for the inner sphere and 180 V for the outer sphere.
1. Charge on the Inner Sphere
Using the capacitance we calculated earlier, we can find the charge \( Q_{inner} \) on the inner sphere:
Q_{inner} = C \cdot V_{inner} = (5.57 \times 10^{-12}) \cdot (90) \approx 5.013 \times 10^{-10} \, C \, (or \, 501.3 \, nC)
2. Charge on the Outer Shell
For the outer shell, since it is grounded, the charge \( Q_{outer} \) can be found using the potential difference and the capacitance:
Q_{outer} = C \cdot V_{outer} = (5.57 \times 10^{-12}) \cdot (180) \approx 1.003 \times 10^{-9} \, C \, (or \, 1003 \, nC)
Summary of Results
- Potential difference between the two conductors: approximately 899.64 V
- Capacitance of the system: approximately 5.57 pF
- Charge on the inner sphere when at 90 V: approximately 501.3 nC
- Charge on the outer shell when at 180 V: approximately 1003 nC
This analysis provides a comprehensive understanding of the electrostatic behavior of the system, illustrating the relationships between charge, potential, and capacitance in spherical conductors.
